Mathematical Problem - Geometry Involved

[Duckers]

I found this puzzle somewhere and I thought it would be fun to share it with you guys.

Imagine any triangle, divide each side into 3 equal parts. Now draw a line from each vertex to the right-most divider.

In terms of the original triangle, what is the area of the smaller triangle in the middle?

Picture: click here - Hope it works now

 

[A5]

...

 

[GoldenBear]

That crosswinds logo isn't a triangle.

 

[GoWaldo]

good point

 

[Duckers]

I just fixed the link

 

[Buddhist]

still dead.

-M.T.O

 

[ace31216]

Link still in progress? Not working.

 

[Time2Kill]

The triangle in the middle is 37.7964473009436666666666666666666666% of the original size of the triangle

it is a 3:1.13389341902831 ratio

 

[jpsj82]

Copy and paste this in your browser to make it work.

http://www.crosswinds.net/~devicecontroller/triangle.bmp

 

[Time2Kill]

You have to right click on the link and click save as to get the pic

 

[Duckers]

Actually A5's link is the one that is screwed up.

His link directs you to: http://www.crosswinds.net/~devicecontroller/triangle.bmp

Mine directs you to:
http://www.crosswinds.net/~devicecontroller/triangle.gif

 

[Time2Kill]

So Duckers, was my answer right?

 

[Duckers]

Time2Kill,

Show work or you will not receive full credit

 

[Mister T]

Duckers,

Its been over 3 years since I have touched the the law of sines....

Anyway, my answer is .368688946 of the original triangle...

This is derived from the following:

sin 60 / sin 100 - sin 20 /(3*sin60) - sin 100 /(3sin 60)

I might have made an error somewhere as I scribbled stuff over a couple pages...

 

[Time2Kill]

I just drew it up in an autocad program, made the sides of the traingle 3 inches long, so each divided section was one inch.

Then I connected all the points, and measured a side of the small triangle, which was 1.13389341902831 inches long. So 3/1.13389341902831 is 37.7964473009436666666666666666666666% of the original size or a 3:1.13389341902831 ratio

 

[Time2Kill]

I just drew it up in an autocad program, made the sides of the traingle 3 inches long, so each divided section was one inch.

Then I connected all the points, and measured a side of the small triangle, which was 1.13389341902831 inches long. So 3/1.13389341902831 is 37.7964473009436666666666666666666666% of the original size or a 3:1.13389341902831 ratio

 

[josphstalinator]

I just drew it up in an autocad program, made the sides of the traingle 3 inches long, so each divided section was one inch.

Then I connected all the points, and measured a side of the small triangle, which was 1.13389341902831 inches long. So 3/1.13389341902831 is 37.7964473009436666666666666666666666% of the original size or a 3:1.13389341902831 ratio

the question states to solve the problem for "any triangle"

you solved the problem for an equilateral triangle. the correct answer should be in the terms of a, b, and c where a, b, and c are the lenght of the sides of the triangle

 

[josphstalinator]

deriving the answer is easy but tedious. you just have to use the law of cosines a bunch of times.

 

[A5]

Ok, my link USED to work, but I'll just go delete it now

 

[Time2Kill]

<< I just drew it up in an autocad program, made the sides of the traingle 3 inches long, so each divided section was one inch.

Then I connected all the points, and measured a side of the small triangle, which was 1.13389341902831 inches long. So 3/1.13389341902831 is 37.7964473009436666666666666666666666% of the original size or a 3:1.13389341902831 ratio

the question states to solve the problem for &quot;any triangle&quot;

you solved the problem for an equilateral triangle. the correct answer should be in the terms of a, b, and c where a, b, and c are the lenght of the sides of the triangle >>



But it doesnt say for the sides...the problem states:
In terms of the original triangle, what is the area of the smaller triangle in the middle?